[Haskell-cafe] Why is it so different between 6.12.1 and
6.10.4_1 ?
zaxis
z_axis at 163.com
Sat Mar 27 02:33:48 EDT 2010
>>Why do you bother with the interior definition of f in there?
Because i want to try a C code style not layout style without `do` syntax
sugar .
Yusaku Hashimoto wrote:
>
>>> fac n = let { f = foldr (*) 1 [1..n] } in f
>>
>> Why do you bother with the interior definition of f in there?
>>
>> fac = product . enumFromTo 1
>
> let fac = do is_zero <- (==0); if is_zero then return 1 else liftM2
> (*) id (fac . pred)
>
> -nwn
>
> On Sat, Mar 27, 2010 at 9:59 AM, Ivan Lazar Miljenovic
> <ivan.miljenovic at gmail.com> wrote:
>> zaxis <z_axis at 163.com> writes:
>>> In 6.10.4_1 under freebsd
>>>> let f x y z = x + y + z
>>> *Money> :t f
>>> f :: (Num a) => a -> a -> a -> a
>>>
>>>> :t (>>=) . f
>>> (>>=) . f :: (Monad ((->) a), Num a) => a -> ((a -> a) -> a -> b) -> a
>>> -> b
>>>> ((>>=) . f) 1 (\f x -> f x) 2
>>>
>>> <interactive>:1:1:
>>> No instance for (Monad ((->) a))
>>> arising from a use of `>>=' at <interactive>:1:1-5
>>> Possible fix: add an instance declaration for (Monad ((->) a))
>>> In the first argument of `(.)', namely `(>>=)'
>>> In the expression: ((>>=) . f) 1 (\ f x -> f x) 2
>>> In the definition of `it': it = ((>>=) . f) 1 (\ f x -> f x) 2
>>>
>>
>> Some definitions and exports got changed, so in 6.12 the (-> a) Monad
>> instance is exported whereas in 6.10 it isn't.
>>
>>> fac n = let { f = foldr (*) 1 [1..n] } in f
>>
>> Why do you bother with the interior definition of f in there?
>>
>> fac = product . enumFromTo 1
>>
>> --
>> Ivan Lazar Miljenovic
>> Ivan.Miljenovic at gmail.com
>> IvanMiljenovic.wordpress.com
>> _______________________________________________
>> Haskell-Cafe mailing list
>> Haskell-Cafe at haskell.org
>> http://www.haskell.org/mailman/listinfo/haskell-cafe
>>
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-----
fac n = let { f = foldr (*) 1 [1..n] } in f
--
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